Intersection number and cup product

It is known that for a closed oriented smooth manifold $$M$$, if A and B are oriented submanifolds of $$M$$, and if A and B intersect transversely, then the Poincare dual of A ∩ B is the cup product of the Poincare duals of A and B(see Ref 1).

When applying to the case where $$M$$ is a 4 dimensional(also closed, oriented, smooth), and $$A$$ and $$B$$ are two dimensional (oriented) submanifolds, we can have

\begin{align} ([A]^* \cup [B]^*)\cap M=([A \cap B]^*\cap M \end{align} where $$[A]^*,[B]^*$$ denote the Poincare dual of $$A, B$$, and similarly for $$[A\cap B]^*$$ is the Poincare dual to the intersection of A and B, namely $$A \cap B$$, which in general is a zero dimensional submanifold of $$M$$.

In understanding the 4 dimensional case, I meet some basic problems:

(1) What is the definition of Poincare dual of $$A$$ if $$A$$ is not a closed submanifold of $$M$$? For example, if $$A$$ is the seifert surface of a line in $$S^4$$, what is the Poincare dual of this seifert surface?

(2) If I want to calculate the selfintersection number of a open oriented 2-dimensional submanifold $$A$$, it seems that I need to take two submanifolds $$A_1$$ and $$A_2$$ which are two representatives of the cohomology class [A] and also intersect transversely, and then apply the above formula \begin{align} ([A_1]^*\cup [A_2]^*)\cap M=[A_1\cap A_2]^*\cap M \end{align} Now if I consider $$M=S^4$$, $$A$$ is the Seifert surface of a line(an embedded one dimensional submanifold) in $$S^4$$, is it possible that the selfintersection number equal to one? Why, or why not?

I am not very sure whether I have made some mistake in describing my puzzles. If you guys have any idea or suggestion or references, please just post, and I am really grateful!

1 Answer

(1) If $$\Sigma$$ is a submanifold with boundary in $$M$$, so that $$\Sigma \cap \partial M = \partial \Sigma$$, then $$\Sigma$$ has a fundamental class in relative homology which you may push forward and dualizing to obtain a class in $$H^2(M)$$. I don't know what you would mean by the Poincare dual of something whose boundary does whatever it feels. I doubt it's what you want.

(2) If $$\Sigma$$ and $$\Sigma'$$ are submanifolds of complementary dimension so that $$\partial \Sigma \cap \Sigma' = \varnothing$$ and similarly $$\Sigma \cap \partial \Sigma' = \varnothing$$, you may still define an intersection product (make them transverse without moving the boundary). In cohomological language (assuming the ambient manifold is closed so I don't need to say "compactly supported cohomology"), you are taking the cup product of cycles in $$H^2(M, \Sigma)$$ and $$H^2(M, \Sigma')$$ to obtain a class in $$H^4(M, \Sigma \cup \Sigma') \cong \Bbb Z$$, the last isomorphism determined by an orientation on $$M$$. These intersection numbers will not be invariant under arbitrary homotopy. Instead, one is allowed to homotope $$f_t: \Sigma \to M$$ under the rule that $$\partial \Sigma$$ stays fixed, and $$f_t(\Sigma) \cap \partial \Sigma' = \varnothing$$ for all $$t \in [0,1]$$. Similarly with $$\Sigma'$$. If you allow more general homotopies the result is unlikely to be invariant.

If $$\Sigma \cap \Sigma'$$ is nonempty then abandon all hope. There is no invariant notion of "intersection number" anymore, since to undo this intersection point you need to push the boundaries off of one another. This breaks the grand rule of the previous paragraph: you can see in practice this is not a well-defined notion.

In particular, no self-intersections.

On a related note, if $$M$$ is noncompact you can define intersection numbers of properly embedded submanifolds, which are invariant under proper isotopy of each. If the pieces have boundary, again you need to fix a neighborhood if that boundary.

• Thanks. Actually I don't assume $\Sigma\cap \partial M=\partial \Sigma$. In fact, I assume $M$ to be closed. I see in some reference(see 1) where Corollary 2.2 says for a compact s-dimensional oriented submanifold $S$(not necessary closed) of the smooth, finite type oriented manifold $M$ of dimension m, we can find a tubular neighborhood $N$ of $S$ in $M$, and define the Poincare dual of $S$ in $N$, then promote it to be the Poincare dual of $S$ in M by some extension. Actually I don't understand much about this. – wln Jan 1 at 4:11
• (2) Sorry, what do you mean by complementary boundary? And why do you say "If Σ∩Σ′ is nonempty then abandon all hope. There is no invariant notion of "intersection number" anymore." I do not understand this, could you say it in more details. – wln Jan 1 at 4:22
• @wln Nowhere in that text is M or S allowed to have boundary, though the author doesn't state it explicitly. (Much does not make sense if they did.) Whenever the author says "compact manifold" they mean compact without boundary, and whenever they say "closed submanifold" they mean S is closed as a subset of M (this is equivalent to what I call being a "properly embedded submanifold). – user98602 Jan 1 at 4:23
• That should have said "complientary dimension". Fixed now. As for (2), already draw what happens if you try to take intersection numbers of [0,2] on the x-axis with the unit circle, as you move the arc outside the unit circle. The intersection numbers changed because you allowed one manifold to move over the boundary of the other. – user98602 Jan 1 at 4:25
• There were a few typos in the response (2) concerning what the boundary is allowed to intersect or not. Sorry about that. It's fixed now. – user98602 Jan 1 at 4:30